Exercise · ZW-ISOLATED-SINGULARITY-004 · ★★☆ Intermediate
The singularity of \(\frac{\cos z-1}{z^2}\) at \(z=0\) is removable. What value must \(f(0)\) take to remove it?
Choose one option.
Show answer
- C. \(-\frac12\)
Explanation
\(\cos z-1=-\frac{z^2}{2}+\frac{z^4}{24}-\cdots\), so the quotient tends to \(-\frac12\). That value makes \(f\) holomorphic at \(0\).
Review the concept: Isolated singularity
An isolated singularity of a function \(f\) is a point \(z_0\) where \(f\) is not holomorphic but is holomorphic on a punctured disc \(0<|z-z_0|<r\) around it.
More practice problems: isolated singularity
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- ZW-ISOLATED-SINGULARITY-008 Classify the singularity of (z)/(sin z) at z=π.
- ZW-ISOLATED-SINGULARITY-009 Classify the singularity of zsin 1z at z=0.